Time Dilation Light Clock Example

[Drakkith]

I'm missing something here on the light clock example from http://en.wikipedia.org/wiki/Time_dilation.

I understand the math on the picture, but I'm missing how that applies to time dilation.
Why does the light pulse take on a diagonal path for the moving observer? Is the light pulse still the same pulse that bouncing between the mirrors, because the picture confuses me and makes it look like one of the mirrors is moving, in which case I could easily understand why the light takes the longer path.

 

[bcrowell]

Both mirrors are moving, in the frame of the second observer. They only show the mirrors at certain selected times, when the beam of light is at that particular mirror.

 

[Drakkith]

Both mirrors are moving, in the frame of the second observer. They only show the mirrors at certain selected times, when the beam of light is at that particular mirror.

How does that relate to time dilation then? If they are moving, then the light emitted can't simply travel in a straight line perpindicular to both mirrors, or it would end up being left behind correct? If so, the light that is being bounced back and forth is moving diagonally and following that longer path. What does that have to do with time dilation though? The clock only runs slower because light has to take a longer path right?

 

[bcrowell]

How does that relate to time dilation then? If they are moving, then the light emitted can't simply travel in a straight line perpindicular to both mirrors, or it would end up being left behind correct? If so, the light that is being bounced back and forth is moving diagonally and following that longer path. What does that have to do with time dilation though? The clock only runs slower because light has to take a longer path right?

You're seeing the same path of the light, just drawn in two different frames. In one frame there's a 90-degree angle between the path of the light and the surface of the mirror; in the other frame it's not 90 degrees. The same would be true if it was a ping-pong ball and this was Galilean relativity.

The only place where SR comes in is that the speed of the beam of light has to be the same in both frames, even though the distances are unequal.

 

[Drakkith]

From the frame of the moving observer, does it look like both mirrors AND the light are moving? Is that why the lights path looks diagonal?

 

[bcrowell]

From the frame of the moving observer, does it look like both mirrors AND the light are moving? Is that why the lights path looks diagonal?

Right!

 

[Drakkith]

Right!

So how does that relate to time dilation? Something to do with the speed of light being constant no matter what frame you are in?

 

[bcrowell]

So how does that relate to time dilation? Something to do with the speed of light being constant no matter what frame you are in?

Right. The speed of the beam of light has to be the same in both frames, even though the distance is greater in the second frame. If it has to travel a greater distance in the second frame, the time must be greater as well. That means that the time is greater in the second frame than in the first.

 

[Drakkith]

Right. The speed of the beam of light has to be the same in both frames, even though the distance is greater in the second frame. If it has to travel a greater distance in the second frame, the time must be greater as well. That means that the time is greater in the second frame than in the first.

Which means that it takes longer for the light to reach each mirror from the POV of the moving observer?

 

[bcrowell]

Which means that it takes longer for the light to reach each mirror from the POV of the moving observer?

Yes. (Assuming you mean the one moving relative to the mirrors -- each is moving, relative to the other.)

 

[Drakkith]

Yes. (Assuming you mean the one moving relative to the mirrors -- each is moving, relative to the other.)

The ramifications from this are profound! I must notify the world at once!

 

[bobc2]

A four dimensional photon filament zig zags between the 4-D surfaces of a light clock in a rest system. A 4-D red observer and 4-D blue observer are straight lines slanted symmetrically with respect to a rest system. The photon start point is at zero time (t0) for the blue guy (photon departs from the right surface), but the start point is at time t1 for the red guy. The photon next hits the left surface at t4 for the blue guy, but hits that same point on the left surface at t3 for the red guy. The photon next contacts the right surface at t6 for the blue guy and t7 for the red guy, etc.

The blue and red guys cannot agree on the times corresponding to the photon points of contact. However, they both agree that at all times, the photon was moving at the speed of light, c (photon 4-D filament always bisects the angle between the X1 coordinate and the X4 coordinate for both blue and red--the ratio dX1/dX4 is always 1).

 

[chinglu1998]

I'm missing something here on the light clock example from http://en.wikipedia.org/wiki/Time_dilation.

I understand the math on the picture, but I'm missing how that applies to time dilation.
Why does the light pulse take on a diagonal path for the moving observer? Is the light pulse still the same pulse that bouncing between the mirrors, because the picture confuses me and makes it look like one of the mirrors is moving, in which case I could easily understand why the light takes the longer path.

You can do any direction of round trip light travel and conclude time dilation.

That experiment you link is where moving frame had light source. Past the y-axis and a little less, a moving light source bends light beams in the direction of motion travel from view of rest frame. Behind the y-axis and a little less, light bends in the negative direction. That is called light aberration under SR or relativistic doppler.

Can find bending of moving light source in section 7.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

But, if you do not perform two way but only one way, time dilation is false.

This easy to show.

Let r be given.

Now, send light pulse when origins same from rest frame.

Rest frame

t = r/c.

Moving frame.
Length contraction
ct' = r/γ + (-vt').

t' = r/(γ(c+v))

This is not time dilation. But if do 2 way, can prove time dilation.

 

[yuiop]

I'm missing something here on the light clock example from http://en.wikipedia.org/wiki/Time_dilation.

I understand the math on the picture, but I'm missing how that applies to time dilation.
Why does the light pulse take on a diagonal path for the moving observer? Is the light pulse still the same pulse that bouncing between the mirrors, because the picture confuses me and makes it look like one of the mirrors is moving, in which case I could easily understand why the light takes the longer path.

You are right that the Wiki diagram is a little confusing. I thought an animation might be clearer but surprisingly there are not many on youtube. This is the best I could find but it might suffice. http://www.youtube.com/watch?v=XhJitbhsKvI&NR=1

 

[bobc2]

Clock ticks regularly for the red guy since the clock is at rest as far as the red guy is concerned. The blue guy sees the clock moving away from him and ticking irregularly.

Now, sitting back with a bird's eye view of the 4-dimensional universe there is nothing confusing. The blue and red guys simply do not experience the same instantaneous 3-D cross-section views of the 4-dimensional universe. By constructing this type of 4-dimensional universe (photon 4-D filaments always bisect the X1 and X4 coordinates for all 4-dimensional observers), nature has ingeneously assured that the laws of physics work the same for each observer, no matter how his X4 coordinate slants (in any reference system you choose). Nature has woven a fabric of 4-dimensional filaments in a special unique pattern that obeys geometric relationships that are manifes to us as physical laws.

If you regard the universe as 4-dimensional and all objects (including bundles of brain neurons) as 4-dimensional as well, then you might raise the question, "Who or what is doing the moving along the observers 4th dimension at the speed of light?"

 

[grav-universe]

Rest frame

t = r/c.

Moving frame.
Length contraction
ct' = r/γ + (-vt').

t' = r/(γ(c+v))

This is not time dilation. But if do 2 way, can prove time dilation.

This is only comparing what each frame views of their own clocks only, what the rest frame views of the rest frame's own clocks and what the moving frame views of the moving frame's own clocks, so is not the time dilation. The time dilation is what an observer in one frame views of the time passing upon a clock in the other frame as compared to the time that passes upon a clock in the observer's own frame.

Let's place clocks A and B at either end of the length r. According to the rest frame, between the two events of the light pulse leaving clock A and arriving at clock B, we have = tB - tA = Δt = r / c, as you had. From the perspective of the moving frame, with clock B traveling behind clock A, we have c (tB' - tA') = c Δt' = r/γ - v Δt', Δt' = (r/γ) / (c+v), also as you had. But these are just the times that each frame sees passing between the events according to their own clocks.

To find what the moving frame measures passing between the rest frame's clocks in terms of the time dilation involved, we must also include simultaneity of relativity. The moving frame says that with clock B behind clock A, clock B is set a time of tl = r v / c^2 ahead of clock A. According to the moving frame, then, the light pulse leaves clock A when clock A reads TA = 0 and clock B reads TB = r v / c^2. The light pulse travels from clock A to clock B in a time of Δt' = (r/γ) / (c+v) according to the moving frame as you have shown. During this time, the moving frame observes clocks A and B to be time dilating by some factor z, so that when the light pulse reaches B, only z Δt' has passed upon each of the rest frame's clocks according to the moving frame, and the rest frame's clocks now read TA' = z Δt' and TB' = z Δt' + r v / c^2.

The moving frame, then, says that the difference in readings between the rest frame's clocks when the pulse departs clock A and arrives at clock B is TB' - TA = z Δt' + r v / c^2. Observers in all frames must agree with this difference in readings upon the clocks for each event since each clock coincides in the same place as the event. The rest frame says that the difference between the readings upon the rest frame's own clocks for the events is TB - TA = r / c. So since the moving frame must agree with this same difference in readings between the same two clocks for the same two events, then

TB' - TA = tB - tA

z Δt' + r v / c^2 = r / c, where Δt' = (r / γ) / (c + v), so

z (r / γ) / (c + v) + r v / c^2 = r / c, and dividing out r,

z (1 / γ) / (c + v) + v / c^2 = 1 / c

z (1 / γ) / (c + v ) = 1 / c - v / c^2

z = γ (c + v) (1 / c - v / c^2)

= γ (c + v) (c - v) / c^2

= γ (1 - (v / c)^2), and since the length contraction is r / γ = sqrt(1 - (v / c)^2) r,
then γ = 1 / sqrt(1 - (v / c)^2) and

z = (1 - (v / c)^2) / sqrt(1 - (v / c)^2)

= sqrt(1 - (v / c)^2) = 1 / γ

z is the time dilation that the moving observer measures passing upon each of the rest frame's clocks as compared to the moving observer's own clocks, which is 1 / γ as it should be, even with the light pulse traveling only one way.

 

[chinglu1998]

This is only comparing what each frame views of their own clocks only, what the rest frame views of the rest frame's own clocks and what the moving frame views of the moving frame's own clocks, so is not the time dilation. The time dilation is what an observer in one frame views of the time passing upon a clock in the other frame as compared to the time that passes upon a clock in the observer's own frame.

Let's place clocks A and B at either end of the length r. According to the rest frame, between the two events of the light pulse leaving clock A and arriving at clock B, we have = tB - tA = Δt = r / c, as you had. From the perspective of the moving frame, with clock B traveling behind clock A, we have c (tB' - tA') = c Δt' = r/γ - v Δt', Δt' = (r/γ) / (c+v), also as you had. But these are just the times that each frame sees passing between the events according to their own clocks.

To find what the moving frame measures passing between the rest frame's clocks in terms of the time dilation involved, we must also include simultaneity of relativity. The moving frame says that with clock B behind clock A, clock B is set a time of tl = r v / c^2 ahead of clock A. According to the moving frame, then, the light pulse leaves clock A when clock A reads TA = 0 and clock B reads TB = r v / c^2. The light pulse travels from clock A to clock B in a time of Δt' = (r/γ) / (c+v) according to the moving frame as you have shown. During this time, the moving frame observes clocks A and B to be time dilating by some factor z, so that when the light pulse reaches B, only z Δt' has passed upon each of the rest frame's clocks according to the moving frame, and the rest frame's clocks now read TA' = z Δt' and TB' = z Δt' + r v / c^2.

The moving frame, then, says that the difference in readings between the rest frame's clocks when the pulse departs clock A and arrives at clock B is TB' - TA = z Δt' + r v / c^2. Observers in all frames must agree with this difference in readings upon the clocks for each event since each clock coincides in the same place as the event. The rest frame says that the difference between the readings upon the rest frame's own clocks for the events is TB - TA = r / c. So since the moving frame must agree with this same difference in readings between the same two clocks for the same two events, then

TB' - TA = tB - tA

z Δt' + r v / c^2 = r / c, where Δt' = (r / γ) / (c + v), so

z (r / γ) / (c + v) + r v / c^2 = r / c, and dividing out r,

z (1 / γ) / (c + v) + v / c^2 = 1 / c

z (1 / γ) / (c + v ) = 1 / c - v / c^2

z = γ (c + v) (1 / c - v / c^2)

= γ (c + v) (c - v) / c^2

= γ (1 - (v / c)^2), and since the length contraction is r / γ = sqrt(1 - (v / c)^2) r,
then γ = 1 / sqrt(1 - (v / c)^2) and

z = (1 - (v / c)^2) / sqrt(1 - (v / c)^2)

= sqrt(1 - (v / c)^2) = 1 / γ

z is the time dilation that the moving observer measures passing upon each of the rest frame's clocks as compared to the moving observer's own clocks, which is 1 / γ as it should be, even with the light pulse traveling only one way.

Good, you agree those are the times in the frames for the light to move to r.

It is not the case then that dt/dt' = 1/γ for this case.

Now, let's do your situation but done this way.

If there had been a clock at origin when light pulsed, that clock at moving origin would beat time dilated r/(cγ) for that interval and the other moving observer would be t' = r/(γ(c+v)) as calculated. Note how relativity of simultaneity is true.

But you do it your way, and both moving observers would have r/(cγ) on their clocks. That makes relativity of simultaneity false when light pulse hits r. Now assume light continue on to 2r/c in rest frame. Your way makes both moving clocks at 2r/(cγ). Note how moving clocks remain synchrionized on the next time interval for rest frame. This is not what the relativity of simultaneity states.